Shear of Open Cross-Section Beams - Comcast.net

clanmurderUrban and Civil

Nov 15, 2013 (3 years and 10 months ago)

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AERSP 301

Shear of beams

(Open Cross
-
section)

Jose Palacios


Shear of Open and Closed Section Beams


Megson


Ch. 17


Open Section Beams


Consider only shear loads applied through shear center (no twisting)


Torsion loads must be considered separately



Assumptions


Axial constraints are negligible



Shear stresses normal to beam surface are negligible


Near surface shear stress = 0


Walls are thin



Direct and shear stresses on planes normal to the beam surface are const through
the thickness



Beam is of uniform section


Thickness may vary around c/s but not along the beam



Thin
-
Walled


Neglect higher order terms of t (t
2
, t
3
, …)




Closed Section Beams


Consider both shear and torsion loading


Force equilibrium: General stress, Strain, and Displacement
Relationships


S



the distance measured
around the c/s from some
convenient origin



σ
z



Direct stress
(due to
bending moments or bending
action of shear loads)







Shear stresses due to
shear loads or torsion loads
(for closed section)



σ
s



Hoop stress, usually zero
(non
-
zero due to internal
pressure in closed section
beams)




zs

=

sz

=



shear flow; shear

force per unit length

q =


t


(positive in the direction of s)


Force equilibrium (cont’d)


From force equilibrium considerations in z
-
direction:







Force equilibrium in s
-
direction gives

Stress Strain Relationships


Direct stress:

z

and

s



strains

z

and

s


Shear stress:




strains


(=

zs

=

zs
)



Express strains in terms of displacements of a
point on the c/s wall


v
t

and v
n
: tangential and normal
displacements in xy plane




Not used

(1/r: curvature of wall in x
-
y plane)

Stress Strain Relationships


To obtain the shear strain, consider the element below:










Shear strain:

Center of Twist


Equivalent to pure rotation about some pt. R


(center of twist [for loading such as pure torsion])


For the point N




Origin O of axes
chosen arbitrarily,
and axes undergo
disp.
u, v,


Center of Twist (cont’d)





But

Equivalent to pure rotation about some pt. R

(center of twist [for loading such as pure torsion])

Center of twist cont…


Also from






Comparing Coefficients with:


Position of Center of Twist

Shear of Open Section Beams


The open section beam supports
shear loads S
x

and S
y

such that there
is no twisting of the c/s (i.e. no torsion
loads)



For this, shear loads must pass
through a point in the c/s called the
SHEAR CENTER


Not necessarily on a c/s member


Use the equilibrium eqn.




And obtaining

z

from basic bending theory

(no hoop stresses,

s

= 0)

Shear of Open Section Beams cont…

From:

Shear of Open Section Beams cont…


Integrating with respect to s starting from an origin at an
open edge (q = 0 at s = 0) gives:






For a c/s having an axis of symmetry, I
xy

= 0. Then eq.
for q
s

simplifies to:

Shear sample problem


Determine the shear flow distribution
in the thin
-
walled z
-
section shown
due to shear load
S
y

applied through
its shear center (no torsion).



Where is the shear center?



And the centroid?

t

y

x

h

h/2

1

2

4

3

Shear Flow Distribution (S
x

= 0):



























s
xy
yy
xx
yy
y
s
xy
yy
xx
xy
y
s
yds
t
I
I
I
I
S
xds
t
I
I
I
I
S
q
0
2
0
2
Shear sample problem

8

12

3
3
3
3
th
I
th
I
th
I
xy
yy
xx

































ds
y
x
h
S
yds
t
I
xds
t
I
I
I
I
S
q
s
y
s
yy
s
xy
xy
yy
xx
y
s
)
48
.
6
32
.
10
(
0
3
0
0
2
Bottom Flange: 1
-
2,
y =
-
h/2, x =
-
h/2 + S
1

0
≤ S
1
≤ h/2






















ds
h
S
h
h
S
q
ds
y
x
h
S
q
s
y
s
y
s
))
2
/
(
48
.
6
)
2
/
(
32
.
10
(
)
48
.
6
32
.
10
(
1
0
1
3
12
0
3
Show this: EXAM TYPE PROBLEM

Shear sample problem

h
S
q
)
h
(S
q
S
hS
S
h
S
ds
h
S
h
h
S
q
y
y
s
y
42
.
0
2
/

2
@
0
)
0
(

1
@
)
74
.
1
16
.
5
(
))
2
/
(
48
.
6
)
2
/
(
32
.
10
(
2
1
1
1
1
2
1
3
0
1
3
12
1




















t

y

x

h

h/2

1

2

4

3

Shear sample problem



2
2
2
2
3
2
2
2
0
3
23
42
.
3
42
.
3
42
.
0
)
(
48
.
6
)
(
42
.
3
(
2
s
hs
h
h
S
q
ds
s
h
h
S
q
y
s
y













In web 2
-
3:


y =
-
h/2 + S
2

x = 0 for 0


S
2


h

Symmetric distribution about
C
x

with max value at
S
2

= h/2 (y = 0)

and positive shear flow along the web

Shear Flow
S
2

= 0

h
S
q
y
42
.
0
2

Shear sample problem

In web 3
-
4:


y =h/2

x = S
3

for 0


S
3


h

3
3
0
3
3
34
)
(
24
.
3
)
(
32
.
10
(
3
q
ds
h
s
h
S
q
s
y










Shear Flow Distribution in z
-
section

Calculation of Shear Center


If a shear load passes
through the shear center,
it will produce NO TWIST




M = 0



If c/s has an axis of
symmetry, the shear
center lies on this axis



For cruciform or angle
sections, the shear center
is located at the
intersection of the sides

Sample Problem


Calculate the shear
center of the thin
-
walled
channel shown here:

Sample problem shear center

The shear center (point S) lies on the horizontal
(C
x
)

axis

at some distance
ξ
s

from the web
. If a shear load
S
y

passes through

the shear center it will produce
no twist.


Let’s look at the shear flow distribution due to S
y
:

Since
I
xy

= 0
and
S
x

= 0





S
xx
y
s
ds
ty
I
S
q
0
)
6
1
(
12
3
h
b
th
I
xx


Further:












S
y
s
ds
y
h
b
h
S
q
0
3
6
1
12
Then:

Sample problem shear center

Along the bottom flange 1
-
2,
y =
-
h/2

1
2
0
1
3
12
6
1
6
2
/
6
1
12
1
s
h
b
h
S
ds
h
h
b
h
S
q
y
S
y





















At point 2:
S
1

= b

b
h
b
h
S
q
y









6
1
6
2
2
Along the web 2
-
3 ,
y =
-
h/2 + S
2

Sample problem shear center



b
h
b
h
S
s
s
h
h
b
h
S
q
ds
s
h
h
b
h
S
q
y
y
S
y










































6
1
6
2
2
6
1
12
2
/
6
1
12
2
2
2
2
3
2
0
2
2
3
23
2
At point 3:
S
1

= h

2
2
3
6
1
6
q
b
h
b
h
S
q
y










Along the top flange 3
-

4 ,
y = h/2

Sample problem shear center



b
h
b
h
S
s
h
h
b
h
S
q
ds
h
h
b
h
S
q
y
y
S
y






































6
1
6
2
6
1
12
2
/
6
1
12
2
3
3
3
0
3
3
34
3
At point 4:
S
3

= b

0
4

q
Good Check!

Sample problem shear center


Shear Flow Distribution due to
S
y

The moments due to this shear flow

distribution should be equal to zero about the shear center

2
0
0
23
1
12
2
2
ds
q
ds
h
q
s
b
h




S

ξ
s

S
y

Solve for
ξ
s

to find the shear center location:

s
h
y
b
h
s
y
y
ds
b
h
b
h
S
ds
s
s
h
h
b
h
S
ds
s
h
h
b
h
S





































0
2
2
0
0
2
2
2
2
3
1
1
2
6
1
6
2
2
6
1
12
2
6
1
6
2








h
b
h
b
s
1
3
2